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Chapter 15: Problem 1402

The axis of the parabola \(9 \mathrm{y}^{2}-16 \mathrm{x}-12 \mathrm{y}-57=0\) is (a) \(\mathrm{y}=0\) (b) \(16 \mathrm{x}+61=0\) (c) \(3 \mathrm{y}-2=0\) (d) \(3 \mathrm{y}-61=0\)

Short Answer

Expert verified
The axis of the given parabola is \[\boxed{16x + 61 = 0}\].

Step by step solution

01

Rewrite the given equation into standard form

First, we need to rewrite the given equation \(9y^2-16x-12y-57=0\) into the standard form of a parabolic equation. The standard form of a parabolic equation is either: \(y - k = a(x - h)^2\) (a vertical parabola) or \(x - h = a(y - k)^2\) (a horizontal parabola) Where (h, k) is the vertex of the parabola and a is a constant that controls the shape of the parabola. Rearrange the given equation as: \(9y^2 - 12y = 16x + 57\) Now complete the square for the y terms: \((y - \frac{b}{2a})^2\), where in this case, a = 9 and b = -12.
02

Completing the Square and Rearranging

To complete the square for the y terms, we need to add and subtract \(\left(\frac{-12}{2 \cdot 9}\right)^{2}\). \((y - \frac{-12}{2 \cdot 9})^2 = (-\frac{2}{3})^2 = \frac{4}{9}\) Now rewrite the equation, completing the square for y: \(9\left(y^2 -\frac{4}{3}y + \frac{4}{9}\right) = 16x + 57 + 9\left(\frac{4}{9}\right)\) Then, simplify the equation: \(9 \left(y - \frac{2}{3}\right)^2 = 16x + 61\) Now, we have the standard form of a parabolic equation: \(x - h = a(y - k)^2\) Comparing the given equation with standard form: \(x - (-61/16) = \frac{1}{16} \left(y-\frac{2}{3}\right)^2\)
03

Identify the Axis of Parabola

Our equation is of the form \(x - h = a(y - k)^2\), which represents a horizontal parabola. The axis of this parabola is parallel to the y-axis. From the standard equation, we can see that the axis is a vertical line with the equation: \(x = -\frac{61}{16}\) Comparing this to the given options, it matches option (b), which is: \(16x + 61 = 0\) Therefore, the axis of the given parabola is: \[\boxed{16x + 61 = 0}\]

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